MHB Bullet imbedding in a block on a frictionless surface

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In an inelastic collision, a 10g bullet traveling at 1000 m/s strikes a 100g block at rest, resulting in a combined velocity calculated using conservation of momentum. The formula yields a final velocity of approximately 90.91 m/s for the bullet-block system. Conservation of energy cannot be applied here, as kinetic energy is not conserved in inelastic collisions. The lack of information regarding the forces during the collision further restricts the use of Newton's Second Law. Therefore, conservation of momentum is the only applicable principle for this scenario.
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A 10g bullet with v = 1000 m/s strikes a 100g block which is at rest. What is their combined velocity? Can this be done with conservation of energy? momemtum?

Here we have inelastic collision so by conservation of momentum
\[
10\cdot 1000 + 0 = 110\cdot v_f\Rightarrow v_f = \frac{1000}{11}m/s.
\]
Can this be done with CoE? If so, how?
 
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Re: bullet imbedding in a block on a frictionless surface

dwsmith said:
A 10g bullet with v = 1000 m/s strikes a 100g block which is at rest. What is their combined velocity? Can this be done with conservation of energy? momemtum?

Here we have inelastic collision so by conservation of momentum
\[
10\cdot 1000 + 0 = 110\cdot v_f\Rightarrow v_f = \frac{1000}{11}m/s.
\]
Can this be done with CoE? If so, how?

No, it cannot, because the collision is inelastic. An elastic collision, by definition, is a collision in which kinetic energy is conserved. With an inelastic collision, it is not conserved. Moreover, you're given no information about the force during the collision, so you can't use Newton's Second Law, or a more general conservation of energy approach. Conservation of Momentum is all that's available to you.
 
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